LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l6, l1, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = ___const_50HAT0 ∧ x2 = _i5HAT0 ∧ x3 = _iHAT0 ∧ x4 = _tmpHAT0 ∧ x1 = ___const_50HATpost ∧ x2 = _i5HATpost ∧ x3 = _iHATpost ∧ x4 = _tmpHATpost ∧ _tmpHAT0 = _tmpHATpost ∧ _i5HAT0 = _i5HATpost ∧ ___const_50HAT0 = ___const_50HATpost ∧ _iHATpost = 0 ∧ ___const_50HAT0 ≤ _i5HAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x = _x4 ∧ _x5 = 1 + _x1 ∧ 1 + _x1 ≤ _x
l2	3	l0:	x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x9 = _x13 ∧ _x10 = _x14 ∧ _x8 = _x12
l1	4	l3:	x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x17 = _x21 ∧ _x18 = _x22 ∧ _x16 = _x20
l3	5	l4:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x25 = _x29 ∧ _x26 = _x30 ∧ _x24 = _x28 ∧ _x24 ≤ _x26
l3	6	l1:	x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x33 = _x37 ∧ _x32 = _x36 ∧ _x38 = 1 + _x34 ∧ 1 + _x34 ≤ _x32
l5	7	l2:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x46 = 0 ∧ _x47 = _x47 ∧ _x48 = 0 ∧ _x45 = 0 ∧ _x40 = _x44
l6	8	l5:	x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x1 = _x53 ∧ x2 = _x54 ∧ x3 = _x55 ∧ x4 = _x56 ∧ _x52 = _x56 ∧ _x50 = _x54 ∧ _x51 = _x55 ∧ _x49 = _x53

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4

and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

Here we consider the SCC { l0, l2 }.

2.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

l0:	−1 + x1 − x2
l2:	−1 + x1 − x2

2.1.2 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

l2:	0
l0:	−1

2.1.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

2.2 SCC Subproblem 2/2

Here we consider the SCC { l1, l3 }.

2.2.1 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

l1:	−1 + x1 − x3
l3:	−1 + x1 − x3

2.2.2 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

l1:	0
l3:	−1

2.2.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

Tool configuration

AProVE

version: AProVE Commit ID: unknown
strategy: Statistics for single proof: 100.00 % (8 real / 0 unknown / 0 assumptions / 8 total proof steps)